Statistics: Probability Depth
A-level Statistics' Probability Depth takes the addition rule, conditional probability and tree diagrams into the settings that combine several ideas at once: reading a Venn diagram or two-way table with formal set notation (union, intersection, complement), testing whether two events are genuinely independent or mutually exclusive rather than assuming it, and reversing a multi-branch tree diagram to find the probability an outcome came from a particular branch, given information about the final result (an application of Bayes' theorem, even where the specification does not use that name). The examiner is testing whether a student picks the correct structure for the information given and can justify independence or mutual exclusivity with a calculation rather than a guess.
Method
- Decide the structure from the question. Two overlapping categories described for a group of people or objects: use a Venn diagram or two-way table with set notation P(A), P(B), P(A and B), P(A or B), P(A'). Asked to test independence: check whether P(A and B) = P(A) x P(B). Asked to test mutual exclusivity: check whether P(A and B) = 0. Multiple stages or branches, with the branch itself unknown given the outcome: use the law of total probability, then reverse it.
- Key results in set notation: P(A or B) = P(A) + P(B) - P(A and B) (the addition rule); P(A') = 1 - P(A) (the complement rule); if A and B are mutually exclusive then P(A and B) = 0, so P(A or B) simplifies to P(A) + P(B).
- To fill in a Venn diagram or two-way table, start from any given intersection or overlap region, then use the given row, column or marginal totals to find the remaining regions by subtraction, checking that every row, column and the grand total are consistent with each other.
- For a tree diagram with three or more branches, multiply the probabilities ALONG each branch to get the probability of that whole path, then ADD the probabilities of every path that leads to the outcome of interest, since different paths are mutually exclusive.
- For a reversed conditional probability ('given that the outcome was X, find the probability it came from branch B'), use P(B|X) = P(B and X) / P(X), where P(B and X) is one branch's path probability and P(X) is the sum of every path's probability that gives X (the law of total probability). This reversal is Bayes' theorem.
- To test independence of A and B from a Venn diagram or table, calculate P(A), P(B) and P(A and B) separately and check numerically whether P(A) x P(B) equals P(A and B); never assume independence just because it seems plausible in context.
Worked example
A company buys resistors from three suppliers: Supplier A provides 50% of resistors, Supplier B provides 30%, and Supplier C provides 20%. The probability a resistor is faulty is 0.02 for Supplier A, 0.05 for Supplier B, and 0.03 for Supplier C. A resistor is chosen at random and found to be faulty. Find the probability that it came from Supplier B.
- Consider a tree with first-stage branches A (0.5), B (0.3) and C (0.2), each followed by Faulty / Not faulty branches.
- Multiply along each branch to the Faulty outcome: P(A and F) = 0.5 x 0.02 = 0.01. P(B and F) = 0.3 x 0.05 = 0.015. P(C and F) = 0.2 x 0.03 = 0.006.
- Use the law of total probability to find P(F): add the three path probabilities: P(F) = 0.01 + 0.015 + 0.006 = 0.031.
- Apply the reversal: P(B|F) = P(B and F) / P(F) = 0.015 / 0.031.
- Final answer: P(B|F) = 15/31, which is approximately 0.484 (3 s.f.).
Practice questions
Try each question, then tap to reveal the answer.
Q1For events A and B, explain what P(A and B) = 0 means, and state the term used to describe A and B in this case.Show answer
Answer: It means A and B cannot both happen at the same time (they share no outcomes). A and B are described as mutually exclusive.
Q2In a class of 30 students, 18 study French, 14 study Spanish, and 7 study both. A student is chosen at random. Find the probability that the student studies French or Spanish (or both).Show answer
Answer: P(F or S) = P(F) + P(S) - P(F and S) = 18/30 + 14/30 - 7/30 = 25/30 = 5/6.
Q3For events A and B, P(A) = 0.4, P(B) = 0.35 and P(A or B) = 0.61. Determine whether A and B are independent.Show answer
Answer: P(A and B) = P(A) + P(B) - P(A or B) = 0.4 + 0.35 - 0.61 = 0.14. Since P(A) x P(B) = 0.4 x 0.35 = 0.14 also, and these are equal, A and B ARE independent.
Q4Explain why two events with non-zero probabilities cannot be both mutually exclusive and independent.Show answer
Answer: If mutually exclusive, P(A and B) = 0. If independent, P(A and B) = P(A) x P(B), which is non-zero whenever both P(A) and P(B) are non-zero. These two requirements contradict each other unless one event has probability 0, so two events with genuinely non-zero probabilities cannot satisfy both.
Q5A bag contains 5 red, 3 blue and 2 green counters. A counter is drawn at random and not replaced, then a second counter is drawn. Find the probability that the two counters are different colours.Show answer
Answer: P(same colour) = P(RR)+P(BB)+P(GG) = (5/10)(4/9) + (3/10)(2/9) + (2/10)(1/9) = 20/90+6/90+2/90 = 28/90 = 14/45. P(different) = 1 - 14/45 = 31/45.
Q6A test for a rare medical condition correctly gives a positive result for 98% of people who have the condition, and gives a false positive for 3% of people who do not have the condition. In the population, 1% of people actually have the condition. A person is chosen at random and tests positive. Find the probability they actually have the condition, giving your answer to 3 significant figures.Show answer
Answer: P(positive) = 0.01(0.98) + 0.99(0.03) = 0.0098+0.0297 = 0.0395. P(condition | positive) = 0.0098/0.0395 = 0.248 (3 s.f.).
Q7Two counters are drawn without replacement from a bag containing 4 red and 6 yellow counters. Explain why the probability that the second counter is red depends on the colour of the first counter drawn.Show answer
Answer: Removing the first counter changes both the number of red counters remaining and the total number of counters remaining, so the conditional probability of drawing red second is different depending on which colour was removed first; this is why the second-stage branches of the tree use different probabilities depending on the first outcome.
Exam-style questions
Written in the style of a A Level Maths exam paper, with a full mark scheme.
In a survey of 150 adults, 90 say they regularly read fiction, 65 say they regularly read non-fiction, and 20 say they read neither. (a) Find the number of adults who read both fiction and non-fiction. (3) (b) Find P(reads fiction | reads non-fiction), giving your answer as a fraction in its simplest form. (2) (c) Determine, showing your working, whether 'reads fiction' and 'reads non-fiction' are independent events for this group. (2)
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Three machines, X, Y and Z, produce 45%, 35% and 20% of a factory's daily output respectively. The percentage of defective items is 4% from machine X, 2% from machine Y, and 6% from machine Z. An item is selected at random from the day's output and found to be defective. Find the probability that it was produced by machine Z, giving your answer to 3 significant figures.
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A fair six-sided dice is rolled once. Let A be the event 'the score is a prime number' and B be the event 'the score is at least 4'. (a) List the outcomes in A and in B, and find P(A) and P(B). (2) (b) Determine, showing your working, whether A and B are mutually exclusive. (1) (c) Determine, showing your working, whether A and B are independent. (2)
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Free printable worksheet
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